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  "body": "\n## Statement\n\nLet $L/K$ be a [[field extension|0WQ1A8SHTZ]], let $a \\in L$ be algebraic over $K$, and let $m_a$ be its [[minimal polynomial|0GK1WB9R98]].\nFor every $f \\in K[x]$ with $f(a) = 0$, the polynomial $m_a$ [[divides|04GNK7K4VP]] $f$ in $K[x]$.\n\n## Remarks\n\nThe proof: [[polynomial division|0MES3KS0ZC]] gives $f = q m_a + r$ with $\\deg r \u003c \\deg m_a$. Then $r(a) = 0$, so $r = 0$ by the minimality of $\\deg m_a$.\nEquivalently, the kernel of the evaluation map $K[x] \\to L$, $f \\mapsto f(a)$, is the ideal $(m_a)$.\nIf $a$ is a root of $f$ and $f$ is monic and irreducible over $K$, then $f = m_a$.\n\n## Explanations\n\n### L1 human\n\nEvery polynomial equation that the element satisfies comes from the minimal polynomial.\nIt is a multiple of the minimal polynomial.\n",
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  "msc": [
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  "name": "The minimal polynomial divides every polynomial vanishing at the element",
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